Orientations & Covering Maps

Proof.

A covering map is a local diffeomorphism, so an orientation of \(M\) pulls back to an orientation of \(E\). Hence \(E\) is orientable.

Proof.

Let \(\mathcal{O}_E\) denote the given orientation on \(E\).

Suppose first that \(M\) is orientable, and let \(q\) be an arbitrary point of \(E\). Because \(M\), the image of the connected manifold \(E\) under the surjection \(\pi\), is connected and orientable, it has exactly two orientations, and one of them, call it \(\mathcal{O}_M\), has the property that \(d\pi_q : T_q E \to T_{\pi(q)} M\) is orientation-preserving. The pullback orientation \(\pi^{*}\mathcal{O}_M\) then agrees with the given orientation at \(q\), so by connectedness it equals \(\mathcal{O}_E\) everywhere. Now let \(\varphi \in \mathrm{Aut}_\pi(E)\). The identity \(\pi \circ \varphi = \pi\) gives \[ \varphi^{*}\mathcal{O}_E = \varphi^{*}\bigl(\pi^{*}\mathcal{O}_M\bigr) = (\pi \circ \varphi)^{*}\mathcal{O}_M = \pi^{*}\mathcal{O}_M = \mathcal{O}_E, \] so \(\varphi\) is orientation-preserving. Hence the action of \(\mathrm{Aut}_\pi(E)\) preserves orientation.

Conversely, suppose the action of \(\mathrm{Aut}_\pi(E)\) is orientation-preserving, and let \(p \in M\). Choose a connected evenly covered neighborhood \(U\) of \(p\). There is a smooth local section \(\sigma : U \to E\), which induces the pullback orientation \(\sigma^{*}\mathcal{O}_E\) on \(U\). Suppose \(\sigma_1 : U \to E\) is any other smooth local section over \(U\). Because \(\pi\) is a normal cover, \(\mathrm{Aut}_\pi(E)\) acts transitively on each fiber, so there is a deck transformation \(\varphi\) with \(\sigma_1(p) = \varphi\bigl(\sigma(p)\bigr)\). Then \(\varphi \circ \sigma\) is a local section agreeing with \(\sigma_1\) at \(p\). Both are lifts of the inclusion \(U \hookrightarrow M\), and \(U\) is connected, so unique lifting gives \(\sigma_1 = \varphi \circ \sigma\) on all of \(U\). Because \(\varphi\) is orientation-preserving, \[ \sigma_1^{*}\mathcal{O}_E = \sigma^{*}\varphi^{*}\mathcal{O}_E = \sigma^{*}\mathcal{O}_E, \] so the orientations induced by \(\sigma\) and \(\sigma_1\) coincide.

We may therefore define an orientation \(\mathcal{O}_M\) on \(M\) by declaring it on each evenly covered open set to be the pullback induced by any local section. If two such sets \(U\) and \(U'\) meet, each point of \(U \cap U'\) has a connected open neighborhood \(W \subseteq U \cap U'\), because \(M\) is locally path-connected. The set \(W\) is again evenly covered, since each sheet over \(U\) meets \(\pi^{-1}(W)\) in a connected set mapped homeomorphically onto \(W\). Local sections over \(U\) and over \(U'\) restrict to local sections over \(W\), so the computation just made shows that the two orientations agree on \(W\). Hence \(\mathcal{O}_M\) is a well-defined orientation, and \(M\) is orientable.

Proof.

The quotient map \(q : S^n \to \mathbb{RP}^n\) identifying antipodal points is a two-sheeted smooth covering map, one of the model examples of smooth coverings. The antipodal map \(\alpha(x) = -x\) is a diffeomorphism of \(S^n\) with \(q \circ \alpha = q\), so it is a deck transformation. Conversely, a deck transformation \(\varphi\) sends a chosen point \(x_0\) to \(x_0\) or to \(-x_0\), and \(\varphi\), the identity, and \(\alpha\) are all lifts of \(q\) through \(q\). Since \(S^n\) is connected for \(n \geq 1\), unique lifting gives \(\varphi = \mathrm{Id}\) or \(\varphi = \alpha\). Thus \(\mathrm{Aut}_q(S^n) = \{\mathrm{Id}, \alpha\}\), which acts transitively on each fiber \(\{x, -x\}\), so \(q\) is a normal covering. Give \(S^n\) its standard orientation, determined by the outward unit normal \(N_x = x\). An ordered basis \((E_1, \dots, E_n)\) of \(T_x S^n\) is positively oriented exactly when \((x, E_1, \dots, E_n)\) is positively oriented in \(\mathbb{R}^{n+1}\). The antipodal map has differential \(-\,\mathrm{Id}\) on \(\mathbb{R}^{n+1}\), so it sends such a basis to \((-E_1, \dots, -E_n)\) at the point \(-x\), where the outward normal is \(N_{-x} = -x\). Comparing orientations, \[ \bigl(N_{-x}, -E_1, \dots, -E_n\bigr) = \bigl(-x, -E_1, \dots, -E_n\bigr) = (-1)^{n+1}\bigl(x, E_1, \dots, E_n\bigr) \] as a relation between signed bases of \(\mathbb{R}^{n+1}\), since negating all \(n+1\) vectors contributes the factor \((-1)^{n+1}\).

Thus \(\alpha\) carries the induced orientation of \(T_x S^n\) to \((-1)^{n+1}\) times the induced orientation of \(T_{-x} S^n\), so \(\alpha\) is orientation-preserving precisely when \((-1)^{n+1} = 1\), that is, when \(n\) is odd. By the criterion for the base of a normal cover, \(\mathbb{RP}^n\) is orientable if and only if the deck action is orientation-preserving, which happens if and only if \(n\) is odd.

Proof.

Realize the open Möbius band as the quotient of \(\mathbb{R}^2\) by the smooth, free, and properly discontinuous action of \(\mathbb{Z}\) given by \[ n \cdot (x, y) = \bigl(x + n,\, (-1)^n y\bigr). \] The quotient map \(q : \mathbb{R}^2 \to E\) onto the quotient \(E\) is a smooth normal covering, and the deck group is the image of this \(\mathbb{Z}\)-action. We take for granted here that the quotient of a connected smooth manifold, with or without boundary, by a smooth, free, and properly discontinuous action of a discrete group is a smooth manifold and that the quotient map is a smooth normal covering whose deck group is the acting group. This also covers the restriction of \(q\) to the strip \(\mathbb{R} \times [-r, r]\), for a fixed \(r \gt 0\), used below.

Give \(\mathbb{R}^2\) its standard orientation, with orientation form \(dx \wedge dy\). The generator \((x, y) \mapsto (x + 1, -y)\) pulls this form back to \(dx \wedge d(-y) = -\,dx \wedge dy\), so it reverses orientation. The deck action is therefore not orientation-preserving, and by the criterion for the base of a normal cover, \(E\) is not orientable. The closed Möbius band, being the image of a strip \(\mathbb{R} \times [-r, r]\) under the same covering, fails to be orientable by the identical argument applied to the restricted cover.

Proof.

Let \(W\) be a component of \(N\). Because \(N\) is locally path-connected, \(W\) is open in \(N\), and as an open subset it is itself locally path-connected. The base \(M\) is also locally path-connected, because every point of \(M\) has an evenly covered neighborhood \(U\), and \(U\) is homeomorphic to a sheet over it, which is a component of the locally path-connected open set \(\pi^{-1}(U)\) and hence an open, locally path-connected subset of \(N\). Thus \(M\), connected by hypothesis, is also locally path-connected, as the definition of a covering map requires of its base.

Let \(U \subseteq M\) be evenly covered. Each component of \(\pi^{-1}(U)\) is connected and therefore lies in a single component of \(N\). Hence \((\pi|_W)^{-1}(U) = \pi^{-1}(U) \cap W\) is either empty or a nonempty union of components of \(\pi^{-1}(U)\), each mapped homeomorphically onto \(U\). In particular, every point of \(\pi(W)\) has a neighborhood that is evenly covered by \(\pi|_W\).

It remains to show that \(\pi|_W\) is surjective. Each point of \(N\) lies in an open sheet that \(\pi\) maps homeomorphically onto an open subset of \(M\), so \(\pi\) is an open map and \(\pi(W)\) is open. If \(p \in M \setminus \pi(W)\) and \(U\) is an evenly covered neighborhood of \(p\), then \(\pi^{-1}(U) \cap W = \varnothing\), since otherwise it would contain a whole sheet over \(U\) and hence a point over \(p\). Thus \(U \subseteq M \setminus \pi(W)\), and \(\pi(W)\) is closed. Since \(W\) is nonempty and \(M\) is connected, \(\pi(W) = M\).

Proof.

We first topologize \(\widehat M\). For each open \(U \subseteq M\) and each orientation \(\mathcal{O}\) of \(U\), let \[ \widehat U_{\mathcal{O}} = \bigl\{ (p, \mathcal{O}_p) : p \in U \bigr\}, \] where \(\mathcal{O}_p\) is the orientation of \(T_pM\) determined by \(\mathcal{O}\). Every point of \(M\) has an orientable open neighborhood. The domain of a smooth chart \((U, \varphi)\) is oriented by pulling back the standard orientation of \(\mathbb{R}^n\) or \(\mathbb{H}^n\), where \(n = \dim M\), through \(\varphi\), and an open subset of an oriented open set inherits the restricted orientation. Given \((p, \mathcal{O}_p) \in \widehat M\), choose such a \(U\) with an orientation \(\mathcal{O}\), replaced by its opposite if necessary so that it determines \(\mathcal{O}_p\) at \(p\). Then \((p, \mathcal{O}_p) \in \widehat U_{\mathcal{O}}\), so these sets cover \(\widehat M\). If \((p, \mathcal{O}_p)\) lies in \(\widehat U_{\mathcal{O}} \cap \widehat{U'}_{\mathcal{O}'}\), let \(V\) be the component of \(U \cap U'\) containing \(p\), which is open because \(M\) is locally path-connected. The restrictions \(\mathcal{O}|_V\) and \(\mathcal{O}'|_V\) agree at \(p\), hence agree on all of \(V\), so \(\widehat V_{\mathcal{O}|_V}\) is contained in the intersection and contains \((p, \mathcal{O}_p)\). The sets \(\widehat U_{\mathcal{O}}\) therefore form a basis for a topology on \(\widehat M\).

The map \(\widehat\pi\) carries each basis set \(\widehat U_{\mathcal{O}}\) bijectively onto \(U\). It is continuous, because for open \(V \subseteq M\) the set \(\widehat\pi^{-1}(V) \cap \widehat U_{\mathcal{O}}\) is the basis set \(\widehat{(U \cap V)}_{\mathcal{O}|_{U \cap V}}\), so \(\widehat\pi^{-1}(V)\) is a union of basis sets. It is open, because it carries each basis set onto an open subset of \(M\). Hence each restriction \(\widehat U_{\mathcal{O}} \to U\) is a continuous open bijection, that is, a homeomorphism, and \(\widehat M\) is locally homeomorphic to \(M\), in particular locally path-connected.

Now let \(U \subseteq M\) be a connected orientable open set with an orientation \(\mathcal{O}\). The fiber over each \(p \in U\) consists of the orientation determined by \(\mathcal{O}\) and its opposite, so \(\widehat\pi^{-1}(U) = \widehat U_{\mathcal{O}} \sqcup \widehat U_{-\mathcal{O}}\), a disjoint union of two open sets each mapped homeomorphically onto \(U\) and hence connected. These two sets are therefore the components of \(\widehat\pi^{-1}(U)\), and \(U\) is evenly covered. Every point of \(M\) has such a neighborhood, for instance the preimage under a chart of a small ball or half-ball, so \(\widehat\pi\) is a generalized covering map. By the lemma on components, its restriction to each component \(W\) of \(\widehat M\) is a covering map onto the connected smooth manifold \(M\). By the propositions on covering spaces of smooth manifolds without and with boundary, each \(W\) is a topological manifold, with boundary if \(M\) has one, and carries a unique smooth structure making \(\widehat\pi|_W\) a smooth covering map. The components are open, and each maps onto \(M\), so each contains a point of every two-point fiber and there are at most two of them. Hence \(\widehat M\) is Hausdorff and second-countable, and the smooth structures on the components combine into one on \(\widehat M\) for which \(\widehat\pi\) is a local diffeomorphism. This proves (a).

Next we orient \(\widehat M\). At \(\hat p = (p, \mathcal{O}_p)\) the differential \(d\widehat\pi_{\hat p}\) is an isomorphism, and we give \(T_{\hat p}\widehat M\) the unique orientation \(\widehat{\mathcal{O}}_{\hat p}\) that it carries to \(\mathcal{O}_p\). On each basis set \(\widehat U_{\mathcal{O}}\), this pointwise orientation is the one that makes \(\widehat\pi|_{\widehat U_{\mathcal{O}}}\) orientation-preserving onto \((U, \mathcal{O})\), which is the orientation pulled back from \((U, \mathcal{O})\) by this local diffeomorphism. It is therefore continuous, and \(\widehat{\mathcal{O}}\) is an orientation of \(\widehat M\).

For (b), every connected orientable open set was shown above to be evenly covered. Conversely, let \(U \subseteq M\) be a connected open set evenly covered by \(\widehat\pi\), and let \(S\) be a component of \(\widehat\pi^{-1}(U)\), which is nonempty by surjectivity. Then \(S\) is open, being a component of an open subset of the locally path-connected space \(\widehat M\), and \(\widehat\pi|_S\) is a bijective local diffeomorphism onto \(U\), hence a diffeomorphism. Its inverse \(\sigma : U \to \widehat M\) is a smooth local section and a local diffeomorphism, so \(\sigma^*\widehat{\mathcal{O}}\) is an orientation of \(U\).

For (c), let \(\mathcal{O}\) be an orientation of an open set \(U \subseteq M\) and put \(\sigma(p) = (p, \mathcal{O}_p)\). Then \(\sigma\) is the inverse of the map \(\widehat\pi|_{\widehat U_{\mathcal{O}}} : \widehat U_{\mathcal{O}} \to U\), which is a bijective local diffeomorphism and hence a diffeomorphism, so \(\sigma\) is smooth. Since \(d\widehat\pi_{\sigma(p)} \circ d\sigma_p\) is the identity and \(d\widehat\pi_{\sigma(p)}\) carries \(\widehat{\mathcal{O}}_{\sigma(p)}\) to \(\mathcal{O}_p\), the differential \(d\sigma_p\) carries \(\mathcal{O}_p\) to \(\widehat{\mathcal{O}}_{\sigma(p)}\), which says \(\sigma^*\widehat{\mathcal{O}} = \mathcal{O}\). An evenly covered open subset of \(M\) with an orientation is a special case.

Proof.

If \(M\) is orientable, let \(\mathcal{O}\) be an orientation of \(M\). Since \(M\) is connected, part (b) of the properties of the orientation covering shows that \(M\) itself is evenly covered, and its proof gives \(\widehat M = \widehat M_{\mathcal{O}} \sqcup \widehat M_{-\mathcal{O}}\), the basis sets for \(U = M\), each homeomorphic to \(M\) and hence connected. Each of these two components is mapped by \(\widehat\pi\) bijectively onto \(M\) as a local diffeomorphism, hence diffeomorphically, and the two components correspond to the two orientations of \(M\).

Now suppose \(M\) is nonorientable. Let \(W\) be a component of \(\widehat M\). By the lemma on components, \(\widehat\pi|_W\) is a covering map onto \(M\), and since it is a local diffeomorphism it is a smooth covering map. Its fibers all have the same cardinality, which is \(1\) or \(2\) because \(W\) is nonempty and the fibers of \(\widehat\pi\) have two points. If the cardinality were \(1\), then \(\widehat\pi|_W\) would be an injective smooth covering map, hence a diffeomorphism, and its inverse would be a smooth global section \(\sigma\) of \(\widehat\pi\). The pullback \(\sigma^*\widehat{\mathcal{O}}\) would then be an orientation of \(M\), contradicting nonorientability.

Therefore the cardinality is \(2\), so \(W\) contains every point of \(\widehat M\) and \(W = \widehat M\). Thus \(\widehat M\) is connected, \(\widehat\pi\) is a covering map by the same lemma, and being a local diffeomorphism it is a two-sheeted smooth covering map.

Proof.

Write \(\mathcal{O}^{\widetilde M}\) for the orientation of \(\widetilde M\), and define \(\varphi : \widetilde M \to \widehat M\) by \[ \varphi(\tilde q) = \bigl(\widetilde\pi(\tilde q),\ d\widetilde\pi_{\tilde q}(\mathcal{O}^{\widetilde M}_{\tilde q})\bigr), \] where the differential \(d\widetilde\pi_{\tilde q}\), an isomorphism because \(\widetilde\pi\) is a local diffeomorphism, carries the orientation of \(T_{\tilde q}\widetilde M\) to an orientation of \(T_{\widetilde\pi(\tilde q)}M\). By construction \(\widehat\pi \circ \varphi = \widetilde\pi\).

To see that \(\varphi\) is smooth, fix \(\tilde q \in \widetilde M\), let \(U\) be an open set evenly covered by \(\widetilde\pi\) that contains \(\widetilde\pi(\tilde q)\), and let \(\widetilde S\) be the sheet over \(U\) containing \(\tilde q\), with inverse diffeomorphism \(s = (\widetilde\pi|_{\widetilde S})^{-1} : U \to \widetilde S\). The pullback \(\mathcal{O}^U = s^*\mathcal{O}^{\widetilde M}\) is an orientation of \(U\), and for \(\tilde x \in \widetilde S\) the differential \(d\widetilde\pi_{\tilde x} = (ds_{\widetilde\pi(\tilde x)})^{-1}\) carries \(\mathcal{O}^{\widetilde M}_{\tilde x}\) to \(\mathcal{O}^U_{\widetilde\pi(\tilde x)}\). Let \(\sigma : U \to \widehat M\), \(\sigma(p) = (p, \mathcal{O}^U_p)\), be the smooth local section of part (c) of the properties of the orientation covering. Then \(\varphi = \sigma \circ \widetilde\pi\) on \(\widetilde S\). Since \(\widetilde\pi|_{\widetilde S}\) is a diffeomorphism onto \(U\) and \(\sigma\) is a diffeomorphism onto the open set \(\widehat U_{\mathcal{O}^U}\), the map \(\varphi\) restricts to a diffeomorphism of \(\widetilde S\) onto an open subset of \(\widehat M\). Hence \(\varphi\) is smooth and a local diffeomorphism.

Next we show that \(\varphi\) is bijective. Let \(A\) be the set of \(p \in M\) over which \(\varphi\) is injective on the two-point fiber \(\widetilde\pi^{-1}(p)\). Let \(U\) be evenly covered by \(\widetilde\pi\). It is connected, being homeomorphic to a sheet, and it has exactly two sheets \(\widetilde S_1\) and \(\widetilde S_2\). Let \(\mathcal{O}_1\) and \(\mathcal{O}_2\) be the orientations of \(U\) they induce as in the preceding paragraph, so that for each \(p \in U\) the two points of \(\widetilde\pi^{-1}(p)\) are sent to \((p, (\mathcal{O}_1)_p)\) and \((p, (\mathcal{O}_2)_p)\). Because \(U\) is connected, \(\mathcal{O}_2\) equals either \(\mathcal{O}_1\) or \(-\mathcal{O}_1\) on all of \(U\). Hence either \(U \subseteq A\) or \(U \cap A = \varnothing\). The sets \(A\) and \(M \setminus A\) are therefore open, and since \(M\) is connected, \(A = M\) or \(A = \varnothing\).

Suppose \(A = \varnothing\). Then over each evenly covered \(U\) as above, \(\varphi(\widetilde M) \cap \widehat\pi^{-1}(U) = \varphi\bigl(\widetilde\pi^{-1}(U)\bigr) = \widehat U_{\mathcal{O}_1}\), and the complement of \(\varphi(\widetilde M)\) in \(\widehat\pi^{-1}(U)\) is \(\widehat U_{-\mathcal{O}_1}\). Since these sets \(U\) cover \(M\), both \(\varphi(\widetilde M)\) and its complement are open in \(\widehat M\), and both are nonempty, the complement because it meets every fiber. This contradicts the connectedness of \(\widehat M\), which holds by the Orientation Covering Theorem because \(M\) is nonorientable. Therefore \(A = M\). Each two-point fiber of \(\widetilde\pi\) is then mapped injectively, hence bijectively, onto the two-point fiber of \(\widehat\pi\) over the same point. Thus \(\varphi\) is bijective, and a bijective local diffeomorphism is a diffeomorphism.

Finally, \(\varphi\) is orientation-preserving. The identity \(\widehat\pi \circ \varphi = \widetilde\pi\) gives \(d\widehat\pi_{\varphi(\tilde q)} \circ d\varphi_{\tilde q} = d\widetilde\pi_{\tilde q}\), so \(d\widehat\pi_{\varphi(\tilde q)}\) carries \(d\varphi_{\tilde q}(\mathcal{O}^{\widetilde M}_{\tilde q})\) to \(d\widetilde\pi_{\tilde q}(\mathcal{O}^{\widetilde M}_{\tilde q})\), the second coordinate of \(\varphi(\tilde q)\). By the definition of the orientation of \(\widehat M\), this says that \(d\varphi_{\tilde q}\) carries \(\mathcal{O}^{\widetilde M}_{\tilde q}\) to the orientation of \(T_{\varphi(\tilde q)}\widehat M\). For uniqueness, let \(\varphi'\) be another orientation-preserving diffeomorphism with \(\widehat\pi \circ \varphi' = \widetilde\pi\). The same computation shows that \(\varphi'(\tilde q)\) has first coordinate \(\widetilde\pi(\tilde q)\) and second coordinate \(d\widetilde\pi_{\tilde q}(\mathcal{O}^{\widetilde M}_{\tilde q})\), so \(\varphi'(\tilde q) = \varphi(\tilde q)\) for every \(\tilde q\).

Proof.

A connected \(0\)-manifold is a single point, which is orientable, so we may assume that \(M\) has positive dimension. Suppose \(M\) is not orientable. By the Orientation Covering Theorem, its orientation covering \(\widehat\pi : \widehat M \to M\) is then a two-sheeted smooth covering map, in particular a covering map with two sheets. Fix \(q \in \widehat M\) and put \(p = \widehat\pi(q)\). By the correspondence between sheets and cosets, the subgroup \(\widehat\pi_*\bigl(\pi_1(\widehat M, q)\bigr)\) has index two in \(\pi_1(M, p)\). Thus nonorientability produces a subgroup of index two. Contrapositively, if the fundamental group of \(M\) has no subgroup of index two, then \(M\) is orientable. A simply connected manifold has trivial fundamental group, whose only subgroup has index one, so it is orientable.